$4(2x^2+1)+3(x^2-2x).\sqrt{x}(2x-1)=2(x^3+5x)$$<=> (x-2)(3x.\sqrt{2x-1}-2x^2+4x-2)=0$$=> x=2$ hoặc $3x.\sqrt{2x-1}=2(x-1)^2<=> (x^2-8x+4)(4x^2-2x+1)=0$
$4(2x^2+1)+3(x^2-2x).\sqrt{x}(2x-1)=2(x^3+5x)$$<=> (x-2)(3x.\sqrt{x}(2x-1)-2x^2+4x-2)=0$cái sau làm chưa ra :D
$4(2x^2+1)+3(x^2-2x).\sqrt{x}(2x-1)=2(x^3+5x)$$<=> (x-2)(3x.\sqrt{2x-1
}-2x^2+4x-2)=0$
$=> x=2$ hoặc
$3x.\sqr
t{2x-1}=2(x-1)^2<=> (x^2-8x+4)(4x^2-2x+1)=0$